Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String in[] = br.readLine().split(" ");
        int n = Integer.parseInt(in[0]);
        in = br.readLine().split(" ");
        double arr[]=new double[n];
        double dp[] = new double[n+1];
        for(int i=0;i<n;i++){
            arr[i]=Double.parseDouble(in[i]);
        }
        dp[0]=1;
        for(int i=0;i<n;i++){
            for(int coin=i+1;coin>=0;coin--){
                dp[coin]= ((coin==0)?0:dp[coin-1]*arr[i])+dp[coin]*(1-arr[i]);
            }
        }
        double ans = 0;
        for(int i=0;i<=n;i++){
            int tail = n-i;
            if(tail<i){
                ans+=dp[i];
            }
        }

        System.out.println(ans);
    }
}